Math Help - Predicates and quantifier ..

1. Predicates and quantifier ..

Can you help me in this question ..

2. please if can give me the steps to solve this question ..

3. Originally Posted by Rozana
Can you help me in this question ..

Apparently this is a multiple choice question, judging from your other posts (which make it clearer).
But still, the wording is a bit puzzling.

By domain of Q, do mean the extension of Q?
If so, then what is the domain under discussion? The set of the first three natural numbers, {0,1,2}?

If so, and we write out the expansion of that existentially quantified formula, it should look like:

~Q(0,0,0) V ~Q(0,0,1) V ~Q(0,0,2)

Then clearly, none of the four will fill the bill.

But of course, I may be completely misreading the question; it won't be the first time.
Sorry, I have to go. I can look back later, but maybe before that someone else will help.

4. Originally Posted by Rozana

In the question 4

I think the answer in none and it should be (f,f,t)

in the statement x=-x is true because 0=(-0)=0

do you agree with me ??

5. I need some hints to solve question 3 ..

and how can I evaluate ~Q(0,0,1)?

6. Originally Posted by Rozana

In the question 4

I think the answer in none and it should be (f,f,t)

in the statement x=-x is true because 0=(-0)=0

do you agree with me ??

sorry the true answer is (t,f,t)

but another question i don't understand how to solve

7. Originally Posted by Rozana
but another question i don't understand how to solve

TFT

Ax x+1 > x
is obviously true, where 'x' ranges over real numbers.

Ax 2x = 3x, is false where 'x' ranges over real numbers, since, e.g. 2*1 does not equal 3*1.

Ex x = -x, is true where 'x' ranges over real numbers, since 0 = -0, so there is at least one real number x for which x = -x is true.

/

a.

You're being asked to translate Ay Q(0 y 0) into a conjunction or disjunction. But the only values allowed for 'y' are 0 or 1.

So for Q(0 y 0) to hold for all allowed values of 'y' is to say that it holds for both 0 and for 1, when 0 or 1, respectively replace 'y'.

So, in these circumstances,

Ay Q(0 y 0)

becomes

Q(0 0 0) /\ Q(0 1 0)

That's all there is to it, really.

8. Hello Rozana

$Q(x,y,z)$ is an example of a propositional function; that is, a function that returns a truth value (i.e. the value True or False) that will depend upon the value(s) of any parameter(s) that are supplied to it.

Here, the value of $Q$ will be determined by the values of three parameters, $x, y$ and $z$. The values that $x, y, z$ can take are given in the question - defined as the domain of $Q(x,y,z)$.

Since $x$ can take $3$ different values $(0,1,2)$; and $y$ can take $2$ different values $(0,1)$; and $z$ can also take $2$ values $(0,1)$, there are $3\times2\times2= 12$ possible propositions that can be represented by $Q(x,y,z)$, each one returning the value True or False. They are:
$Q(0,0,0),\; Q(0,0,1),\; Q(0,1,0),\;Q(0,1,1)$

$Q(1,0,0),\; Q(1,0,1),\; Q(1,1,0),\;Q(1,1,1)$

$Q(2,0,0),\; Q(2,0,1),\; Q(2,1,0),\;Q(2,1,1)$

Without knowing the actual details of what $Q$ is, we can't say what value each of these propositions has. However, we can say, that if (as in your original question) $\exists z,\; \neg Q(0,0,z)$ is a true statement, then for one or both of the possible values of $z$ ( $0$ and $1$), $\neg Q(0,0,z)$ is true. In other words, $\neg Q(0,0,0)$ is true or $\neg Q(0,0,1)$ is true (or both). So:
$\exists z,\; \neg Q(0,0,z)$ can be written as $\neg Q(0,0,0) \lor \neg Q(0,0,1)$
****

Question 3 can be answered in the same way:
$\forall y,\;Q(0,y,0)$
means
For all $y$ (that is, $y = 0$ and $y = 1$), $Q(0,y,0)$ is true.
So
$Q(0,0,0)$ is true and $Q(0,1,0)$ is true.
Therefore:
$\forall y,\;Q(0,y,0)$ can be written as $Q(0,0,0) \land Q(0,1,0)$
****

Question 4

Examine the truth of each of the propositions in turn:
$p:\forall x, \; x+1>x$
$p$ is True. It simply says that if we add $1$ to any real number, the answer is greater than the number we started with.
$q:\forall x, \;2x = 3x$
Clearly $q$ is False. To show this, we only need find one value of $x$ for which $2x \ne 3x$. (That's not very hard, is it?)
$r:\exists x,\;x=-x$
$r$ is True. There is a value of $x$ for which $x = -x$; that is $x = 0$.

So the ordered triple $(p,q,r)$ has the truth value $(T, F, T)$.

Does that clear up the problems?